Theorem bj-dfsbc

Description: Proof of df-sbc when taking bj-df-sb as definition. (Contributed by BJ, 19-Feb-2026) (Proof modification is discouraged.) (New usage is discouraged.)

	Ref	Expression
Assertion	bj-dfsbc	⊢ ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ [ 𝐴 / 𝑥 ] 𝜑 )

Proof

Step	Hyp	Ref	Expression
1		df-clab	⊢ ( 𝑦 ∈ { 𝑥 ∣ 𝜑 } ↔ [ 𝑦 / 𝑥 ] 𝜑 )
2		sb6	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 ↔ ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) )
3	1 2	bitri	⊢ ( 𝑦 ∈ { 𝑥 ∣ 𝜑 } ↔ ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) )
4	3	anbi2i	⊢ ( ( 𝑦 = 𝐴 ∧ 𝑦 ∈ { 𝑥 ∣ 𝜑 } ) ↔ ( 𝑦 = 𝐴 ∧ ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) ) )
5	4	exbii	⊢ ( ∃ 𝑦 ( 𝑦 = 𝐴 ∧ 𝑦 ∈ { 𝑥 ∣ 𝜑 } ) ↔ ∃ 𝑦 ( 𝑦 = 𝐴 ∧ ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) ) )
6		dfclel	⊢ ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ ∃ 𝑦 ( 𝑦 = 𝐴 ∧ 𝑦 ∈ { 𝑥 ∣ 𝜑 } ) )
7		bj-df-sb	⊢ ( [ 𝐴 / 𝑥 ] 𝜑 ↔ ∃ 𝑦 ( 𝑦 = 𝐴 ∧ ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) ) )
8	5 6 7	3bitr4i	⊢ ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ [ 𝐴 / 𝑥 ] 𝜑 )